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STATICS: CE201
Chapter 7
Internal Forces
Notes are prepared based on: Engineering Mechanics, Statics by R. C. Hibbeler, 12E Pearson
Dr M. Touahmia & Dr M. Boukendakdji
Civil Engineering Department, University of Hail
(2012/2013)
7. Internal Forces
________________________________________________________________________________________________________________________________________________
Chapter Objective: Use the method of sections to
determine internal forces in 2-D load cases.
Contents:
7.1
7.2
7.3
Internal Forces Development in Structural Members
Shear and Moment Equations and Diagram
Cables
Chapter 7: Internal Forces
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7.1 Internal Forces Development in Structural
Members

The design of any structural member requires to know
both the external loads acting on the member and the
internal forces acting within the member in order to be
sure the material can resist these loading.
 The concrete supporting a bridge
has fractured: What might have
caused it to do this?

Is it because of the internal forces?

If so, what are they and how can we
design these structures to make
them safer?
Chapter 7: Internal Forces
7.1 Internal Forces Development in Structural
Members

If a coplanar force system acts on a member, then in
general a resultant internal normal force N (acting
perpendicular to the section), shear force V (acting along
the surface), and bending moment M will act at any
cross section along the member.
Chapter 7: Internal Forces
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Steps for Determining Internal Loadings:

Internal Loadings can be determined by using the
method of section. The following example explains the
steps that we should follow to determine the internal
forces acting on the cross section at point C.
1.
Before the member is sectioned, it is first necessary to
determine its support reactions by drawing a FBD of
the entire structure and solving for the unknown
reactions.
Chapter 7: Internal Forces
Steps for Determining Internal Loadings:
3.
Pass an imaginary section a-a perpendicular to the axis
of the beam through point C and separate the beam into
2 segments. The internal loadings acting at C will be
exposed and become external on the FBD of each
segment: Then, decide which resulting section or piece
will be easier to analyze.
Chapter 7: Internal Forces
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Steps for Determining Internal Loadings:
3.
Draw a FBD of the piece of the structure you’ve
decided to analyze. Remember to show the N, V, and
M loads at the “cut” surface.
4.
Apply the Equations of Equilibrium to the FBD (drawn
in step 3) and solve for the unknown internal loads.
F
x
0
F
y
0
M
C
0
Chapter 7: Internal Forces
Sign Convention:
Chapter 7: Internal Forces
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Example 1

Determine the normal force, shear force and bending
moment acting just to the left, point B, and just to the
right, point C, of the 6 kN force on the beam.
Chapter 7: Internal Forces
Solution 1
1.
Support Reactions: Can be determined from the freebody diagram of the beam.
M
D
0
9 kN.m  6 kN6 m  Ay 9 m  0
Ay  5 kN
Chapter 7: Internal Forces
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Solution 1
2.
Free-Body Diagrams: The free-body diagrams of the
left segments AB and AC of the beam are:
3.
Equations of Equilibrium:
Segment AB:
F
x
0
F
y
0
M
B
0
NB  0
5 kN  VB  0
 5 kN3 m  M B  0
VB  5 kN
M B  15 kN.m
Chapter 7: Internal Forces
Solution 1
Segment AC:
F
x
F
y
0
0
M  0
c
NC  0
5 kN  6 kN  VC  0
 5 kN3 m  M C  0
VC  1 kN
M C  15 kN.m
Chapter 7: Internal Forces
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Example 2

Determine the normal force, shear force and bending
moment at point C of the beam.
Chapter 7: Internal Forces
Solution 2

Note: It is not necessary to find the support reactions at
A since segment BC of the beam can be used to
determine the internal loadings at C.
 1.5 m 
wC  1200 N/m
  600 N/m
 3m 

Free-Body Diagram: The distributed load acting on
segment BC can be replaced by its resultant force:
Chapter 7: Internal Forces
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Solution 2

Equations of Equilibrium:
F
x
F
NC  0
0
V  450 N  0
y
M

0
C
C
0
 M  450 N 0.5 m  0
VC  450 N
M  225 N.m
C
C
Note: The negative sign indicates that MC acts in the
opposite sense to that shown on the free body diagram.
Chapter 7: Internal Forces
7.2 Shear and Moment Diagrams

Beams are designed to support loads perpendicular to
their axes.

The design of a beam requires a detailed knowledge of
the variation of the internal shear force V and bending
moment M acting at each point along the axis of the
beam.
Chapter 7: Internal Forces
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7.2 Shear and Moment Diagrams

To construct the shear and moment diagrams, it is
necessary to section the member at an arbitrary point,
located at distance x from the left end.

The variations of V and M as functions of the position x
along the beam’s axis can be obtained using the method
of sections.

The graphical variations of V and M as functions of x
are termed the shear diagram and bending moment
diagram respectively.
Chapter 7: Internal Forces
7.2 Shear and Moment Diagrams

shear diagram and bending moment diagram of the
beam:
Chapter 7: Internal Forces
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Example 3

Draw the shear and moment diagrams for the shaft
shown in the figure below. The support at A is a thrust
bearing and the support at C is a journal bearing.
Chapter 7: Internal Forces
Solution 3

Support Reactions:
M
F
y
0
C
0
 Ay 4 m  5 kN 2 m  0
 5 kN  2.5 kN  C y  0
Ay  2.5 kN
C y  2.5 kN

Shear and Moment Functions: The shaft is sectioned at
an arbitrary distance x from point A, extending within
the segment AB. The FBD of the left segment is:

Applying the equations of equilibrium:
F
y
0
M  0
V  2.5 kN
(1)
 2.5 kNx   M  0
M  2.5x kN.m
(2)
Chapter 7: Internal Forces
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Solution 3

A free body diagram for the left segment of the shaft
extending a distance x within the region BC is shown:

Applying the equilibrium equations yields:
F 0
M  0
y
2.5 kN  5 kN  V  0
V  2.5 kN (3)
M  5 kNx  2m  2.5 kNx   0
M  10  2.5x  kN.m
(4)
Chapter 7: Internal Forces
Solution 3

Shear and Moment Diagrams: When equations (1) to (4)
are plotted within the regions in which they are valid,
the shear and moment diagrams are obtained.

The shear diagram indicates that the
internal shear force is always 2.5 kN
(positive) within segment AB. Just to
the right of point B, the shear force
changes sign and remains at a constant
value of -2.5 kN for segment BC.

The moment diagram starts at zero,
increases linearly to point B at x = 2 m,
where Mmax = 2.5 kN (2 m) = 5 kN.m,
and therefore decreases back to zero.
Chapter 7: Internal Forces
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Example 4

Draw the shear and moment diagrams for the beam
shown in the figure below.
Chapter 7: Internal Forces
Solution 4

Support Reactions: The support reactions are shown on
the beam’s free-body diagram.

Shear and Moment Functions:
F
y
0
9
1 2
x V  0
3

x2 
V   9   kN
3 

M  0
(1)
1 2 x 
x    9x  0
3 3

x3 
M   9 x   kN.m
9 

M
(2)
Chapter 7: Internal Forces
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Solution 3


Shear and Moment Diagrams: The shear and moment
diagrams are obtained by plotting Eqs. (1) (2).
The point of zero shear can be found using Eq. (1):
V 9

x2
 0,
3
x  5.20 m
Maximum moment can be found
using the value of x = 5.2 m in
Eq. (2):

5.23   31.2 kN.m
M max   95.2 
9 

Chapter 7: Internal Forces
QUIZ

Draw the shear and moment diagrams for the beam
shown in the figure below.
Chapter 7: Internal Forces
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QUIZ

Draw the shear and moment diagrams for the beam
shown in the figure below.
Chapter 7: Internal Forces
QUIZ

Draw complete shear force and bending moment
diagrams for the beam below.
Chapter 7: Internal Forces
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QUIZ

diagrams
Draw complete shear force and bending moment
for
the beam shown in the figure.
7.3 Cables

Flexible cables and chains combine strength with
lightness and often are used in structures for support and
to transmit loads from one member to another.

Cables are used to support suspension bridges and
trolley wheels.
Chapter 7: Internal Forces
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7.3 Cables

Two assumptions are considered in the force analysis of
these systems:
The weight of the cable is negligible.
The cable is perfectly flexible and inextensible.
1.
2.
Cable Subjected to Concentrated loads:

When a cable of negligible
weight
supports
several
concentrated loads, the cable
takes the form of several
straight line segments, each of
which is subjected to a constant
tensile.
Chapter 7: Internal Forces
7.3 Cables

The equilibrium analysis is performed by writing down
a sufficient number of equilibrium equations and
equations describing the geometry of the cable to solve
for all the unknowns leading to a description of the
tension in each segment of the cable.

Example: Consider the cable shown in the figure, where
h, L1, L2, L3, P1 and P2 are known:

The problem here is to
determine the nine unknowns:
Tension in each of the three
segments,
the
four
components of reaction at A
and B and the two sags yC and
yD at point C and D.
Chapter 7: Internal Forces
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Example 5

Determine the tension in each segment of the cable
shown in the figure below:
Chapter 7: Internal Forces
Solution 5


Unknowns: There are 4 unknown external reactions (Ax,
Ay, Ex, Ey), 4 unknown cable tensions (one in each cable
segment) and 2 unknown sags (yB and yD).
Consider the free-body diagram for the entire cable:
 F  0,  A  E  0 A  E
x
M
x
E
x
x
x
0
 Ay 18 m   4 kN15 m   15 kN10 m 
 3 kN2 m   0
Ay  12 kN
 F  0, 12kN  4kN  15kN  3kN  E  0
y
y
E  10kN
y
Chapter 7: Internal Forces
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Solution 5

Consider the leftmost segment, which cuts cable BC:
M
0
C
Ax 12 m  12 kN8 m  4 kN5 m  0
Ax  E x  6.33 kN
F
x
F

y
0
TBC cos  BC  6.33 kN  0
0
12 kN  4 kN  TBC sin  BC  0
Thus:
 BC  51.6 
TBC  10.2 kN
Chapter 7: Internal Forces
Solution 5

Proceeding now to analyze the equilibrium of point A,
C, and E in sequence, we have:

Point A:
F
F
x
0
TAB cos  AB  6.33 kN  0
y
0
 TAB sin  AB  12 kN  0
 AB  62.2 

TAB  13.6 kN
Point C:
F
x
F
y
0
TCD cos  CD  12.2 cos 51.6  kN  0
0
TCD sin  CD  10.2 sin 51.6 kN  15 kN  0
 CD  47.9 
TCD  9.44 kN
Chapter 7: Internal Forces
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Solution 5

Point E:
F
F
x
0
6.33 kN  TED cos  ED  0
y
0
10 kN  TED sin  ED  0
 ED  57.7 

TED  11.8 kN
Sags yB and yD:
y B  3 tan 62.2  5.69 m
y D  2 tan 57.7   3.16 m
Chapter 7: Internal Forces
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